3.380 \(\int (c x)^{7/2} (\frac{a}{x^3}+b x^n)^{3/2} \, dx\)

Optimal. Leaf size=122 \[ -\frac{2 a^{3/2} c^4 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{a}}{x^{3/2} \sqrt{\frac{a}{x^3}+b x^n}}\right )}{(n+3) \sqrt{c x}}+\frac{2 a c^2 (c x)^{3/2} \sqrt{\frac{a}{x^3}+b x^n}}{n+3}+\frac{2 (c x)^{9/2} \left (\frac{a}{x^3}+b x^n\right )^{3/2}}{3 c (n+3)} \]

[Out]

(2*a*c^2*(c*x)^(3/2)*Sqrt[a/x^3 + b*x^n])/(3 + n) + (2*(c*x)^(9/2)*(a/x^3 + b*x^n)^(3/2))/(3*c*(3 + n)) - (2*a
^(3/2)*c^4*Sqrt[x]*ArcTanh[Sqrt[a]/(x^(3/2)*Sqrt[a/x^3 + b*x^n])])/((3 + n)*Sqrt[c*x])

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Rubi [A]  time = 0.276027, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2028, 2031, 2029, 206} \[ -\frac{2 a^{3/2} c^4 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{a}}{x^{3/2} \sqrt{\frac{a}{x^3}+b x^n}}\right )}{(n+3) \sqrt{c x}}+\frac{2 a c^2 (c x)^{3/2} \sqrt{\frac{a}{x^3}+b x^n}}{n+3}+\frac{2 (c x)^{9/2} \left (\frac{a}{x^3}+b x^n\right )^{3/2}}{3 c (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(7/2)*(a/x^3 + b*x^n)^(3/2),x]

[Out]

(2*a*c^2*(c*x)^(3/2)*Sqrt[a/x^3 + b*x^n])/(3 + n) + (2*(c*x)^(9/2)*(a/x^3 + b*x^n)^(3/2))/(3*c*(3 + n)) - (2*a
^(3/2)*c^4*Sqrt[x]*ArcTanh[Sqrt[a]/(x^(3/2)*Sqrt[a/x^3 + b*x^n])])/((3 + n)*Sqrt[c*x])

Rule 2028

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*p*(n - j)), x] + Dist[a/c^j, Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c,
j, m, n}, x] && IGtQ[p + 1/2, 0] && NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2031

Int[((c_)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracPar
t[m])/x^FracPart[m], Int[x^m*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && IntegerQ[p + 1/2]
&& NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0]

Rule 2029

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int (c x)^{7/2} \left (\frac{a}{x^3}+b x^n\right )^{3/2} \, dx &=\frac{2 (c x)^{9/2} \left (\frac{a}{x^3}+b x^n\right )^{3/2}}{3 c (3+n)}+\left (a c^3\right ) \int \sqrt{c x} \sqrt{\frac{a}{x^3}+b x^n} \, dx\\ &=\frac{2 a c^2 (c x)^{3/2} \sqrt{\frac{a}{x^3}+b x^n}}{3+n}+\frac{2 (c x)^{9/2} \left (\frac{a}{x^3}+b x^n\right )^{3/2}}{3 c (3+n)}+\left (a^2 c^6\right ) \int \frac{1}{(c x)^{5/2} \sqrt{\frac{a}{x^3}+b x^n}} \, dx\\ &=\frac{2 a c^2 (c x)^{3/2} \sqrt{\frac{a}{x^3}+b x^n}}{3+n}+\frac{2 (c x)^{9/2} \left (\frac{a}{x^3}+b x^n\right )^{3/2}}{3 c (3+n)}+\frac{\left (a^2 c^4 \sqrt{x}\right ) \int \frac{1}{x^{5/2} \sqrt{\frac{a}{x^3}+b x^n}} \, dx}{\sqrt{c x}}\\ &=\frac{2 a c^2 (c x)^{3/2} \sqrt{\frac{a}{x^3}+b x^n}}{3+n}+\frac{2 (c x)^{9/2} \left (\frac{a}{x^3}+b x^n\right )^{3/2}}{3 c (3+n)}-\frac{\left (2 a^2 c^4 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{1}{x^{3/2} \sqrt{\frac{a}{x^3}+b x^n}}\right )}{(3+n) \sqrt{c x}}\\ &=\frac{2 a c^2 (c x)^{3/2} \sqrt{\frac{a}{x^3}+b x^n}}{3+n}+\frac{2 (c x)^{9/2} \left (\frac{a}{x^3}+b x^n\right )^{3/2}}{3 c (3+n)}-\frac{2 a^{3/2} c^4 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{a}}{x^{3/2} \sqrt{\frac{a}{x^3}+b x^n}}\right )}{(3+n) \sqrt{c x}}\\ \end{align*}

Mathematica [A]  time = 0.0839508, size = 100, normalized size = 0.82 \[ \frac{2 c^2 (c x)^{3/2} \sqrt{\frac{a}{x^3}+b x^n} \left (\sqrt{a+b x^{n+3}} \left (4 a+b x^{n+3}\right )-3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^{n+3}}}{\sqrt{a}}\right )\right )}{3 (n+3) \sqrt{a+b x^{n+3}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(7/2)*(a/x^3 + b*x^n)^(3/2),x]

[Out]

(2*c^2*(c*x)^(3/2)*Sqrt[a/x^3 + b*x^n]*(Sqrt[a + b*x^(3 + n)]*(4*a + b*x^(3 + n)) - 3*a^(3/2)*ArcTanh[Sqrt[a +
b*x^(3 + n)]/Sqrt[a]]))/(3*(3 + n)*Sqrt[a + b*x^(3 + n)])

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Maple [F]  time = 0.322, size = 0, normalized size = 0. \begin{align*} \int \left ( cx \right ) ^{{\frac{7}{2}}} \left ({\frac{a}{{x}^{3}}}+b{x}^{n} \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(7/2)*(a/x^3+b*x^n)^(3/2),x)

[Out]

int((c*x)^(7/2)*(a/x^3+b*x^n)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{n} + \frac{a}{x^{3}}\right )}^{\frac{3}{2}} \left (c x\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)*(a/x^3+b*x^n)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^n + a/x^3)^(3/2)*(c*x)^(7/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)*(a/x^3+b*x^n)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(7/2)*(a/x**3+b*x**n)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{n} + \frac{a}{x^{3}}\right )}^{\frac{3}{2}} \left (c x\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)*(a/x^3+b*x^n)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^n + a/x^3)^(3/2)*(c*x)^(7/2), x)